At regular competition Vladik and Valera won a and b candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.

More formally, the guys take turns giving each other one candy more than they received in the previous turn.

This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, theydon’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.

Single line of input data contains two space-separated integers a, b (1 ≤ a, b ≤ 109) — number of Vladik and Valera candies respectively.

Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.

1 1

Valera

7 6

Vladik

Illustration for first test case:

Illustration for second test case:

題目大意：

　　　　　Vladik and Valera 各有a塊和b塊糖，他兩個輪流送給對方糖果，Vladik and Valera 1塊、

　　　　　Valera送給Vladik 2塊、Vladik and Valera 3塊....直到有一方無法送出相應的糖果時結束。

　　　　　輸出對應的人名。

解題思路：暴力模擬即可。

 1 #include <stdio.h>
 2 int main ()
 3 {
 4     int a,b;
 5     while (~scanf("%d%d",&a,&b))
 6     {
 7         for (int i = 1;a>=0&&b>=0; i ++)
 8         {
 9             if (i%2) a-= i;
10             else b -= i;
11         }
12         if (a < 0)
13             printf("Vladik\n");
14         else
15             printf("Valera\n");
16     }
17     return 0;
18 }

Vladik had started reading a complicated book about algorithms containing n pages. To improve understanding of what is written, his friends advised him to read pages in some order given by permutation P = [p1, p2, ..., pn], where pi denotes the number of page that should be read i-th in turn.

Sometimes Vladik’s mom sorted some subsegment of permutation P from position l to position r inclusive, because she loves the order. For every of such sorting Vladik knows number x — what index of page in permutation he should read. He is wondered if the page, which he will read after sorting, has changed. In other words, has px changed? After every sorting Vladik return permutation to initial state, so you can assume that each sorting is independent from each other.

First line contains two space-separated integers n, m (1 ≤ n, m ≤ 104) — length of permutation and number of times Vladik‘s mom sorted some subsegment of the book.

Second line contains n space-separated integers p1, p2, ..., pn (1 ≤ pi ≤ n) — permutation P. Note that elements in permutation are distinct.

Each of the next m lines contains three space-separated integers li, ri, xi (1 ≤ li ≤ xi ≤ ri ≤ n) — left and right borders of sorted subsegment in i-th sorting and position that is interesting to Vladik.

For each mom’s sorting on it’s own line print "Yes", if page which is interesting to Vladik hasn‘t changed, or "No" otherwise.

5 5
5 4 3 2 1
1 5 3
1 3 1
2 4 3
4 4 4
2 5 3

Yes
No
Yes
Yes
No

6 5
1 4 3 2 5 6
2 4 3
1 6 2
4 5 4
1 3 3
2 6 3

Yes
No
Yes
No
Yes

Explanation of first test case:

1. [1, 2, 3, 4, 5] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes".
2. [3, 4, 5, 2, 1] — permutation after sorting, 1-st element has changed, so answer is "No".
3. [5, 2, 3, 4, 1] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes".
4. [5, 4, 3, 2, 1] — permutation after sorting, 4-th element hasn’t changed, so answer is "Yes".
5. [5, 1, 2, 3, 4] — permutation after sorting, 3-rd element has changed, so answer is "No".

題目大意：

　　　　　輸入n個整數，然後有m次詢問。每次詢問輸入三個整數L，R，X，將[L,R]範圍內的整數從小到大排序，

　　　　　判斷第X個整數的位置是否發生變化。

解題思路：

　　　　　下午寫的時候，特意看了下數據範圍，發現不是太大就直接sort了。不過最後被無情的hack了。。。。

　　　　　後來在codeforces群裏看大佬們討論時提到一種思路，對於每一個詢問只用判斷[L,R]中小於第X個整數的個數

　　　　　是否等於X-L（因為 (1 ≤ li ≤ xi ≤ ri ≤ n) 所以只要在[L,R]中有X-L個整數小於Xi排序之後就不會影響Xi的位置）

 1 #include <stdio.h>
 2 int main ()
 3 {
 4     int n,m,i,l,r,x;
 5     int p[10010];
 6     while (~scanf("%d%d",&n,&m))
 7     {
 8         for (i = 1; i <= n; i ++)
 9             scanf("%d",&p[i]);
10         while (m --)
11         {
12             int sum = 0;
13             scanf("%d%d%d",&l,&r,&x);
14             for(i = l; i <= r; i ++)
15                 if (p[i] < p[x])
16                     sum ++;
17             if (x-l == sum)
18                 printf("Yes\n");
19             else
20                 printf("No\n");
21         }
22     }
23     return 0;
24 }

Codeforces Round #416 (Div. 2) A+B